Question

Electrochemistry question? (Voltaic Cell)

A: Which electrode is the cathode in the reaction above?

B: Give the half reaction occurring at each of the electrodes below. Make sure it shows the direction the reaction actually proceeds in the cell above.

C: Write a balanced chemical equation describing the total reaction associated with the electrochemical cell.

D: What is the EMF of the cell above if it is operated under standard state conditions?

Answer 1

I will break it down for you so you can understand how to look at these types of questions.

Explanation:

This is a Galvanic cell. Galvanic cells are spontaneous meaning they have a `-DeltaG^@`. A `-DeltaG^@` just means the reaction occurs without the input of any outside energy source, under standard conditions that is. The standard cell potential of the cell, `E^@`, is always positive in a galvanic cell. Basically, Gibbs and Standard Cell Potential have opposite signs.

You have solid `"Cu"` and `"Zn"` in their respective half cells. Looking at the standard reduction potentials provided, the more positive value means the more likely that species is going to be reduced. A quick look at the table will show you that `"Cu"^(+2)` ions will readily reduce to form solid `"Cu"` than `"Zn"^(+2)` ions will to become `"Zn"` solid.

`color(white)(aaaa)color(magenta)["Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s) color(white)(aaaaa)"E"^@("red") = +0.34]` `("more positive red. potential means reduction will occur")`

`color(white)(aaaa)color(green)["Zn"^(+2)(aq) + 2e^(-)->"Zn"(s) color(white)(aaaaa)"E"^@("red") = -0.76]` `("less positive red. potential means oxidation occurs")`

`color(white)(aaaaaaaaaaaaaaaa)`
`-------------------`

`color(blue)("A: Which electrode is the cathode in the reaction above?")`

• Since we know that`"Cu"^(+2)` ions are more likely to be reduced to become `"Cu"(s)`, the `color(red)("Cu electrode is labeled as the cathode")` while the `color(red)("Zn electrode is labeled as the anode")`, where oxidation occurs. (remember the memory aid AN OX, RED CAT)

`color(blue)("B: Give the half reaction occurring at each of the electrodes below.")`

`color(blue)("Make sure it shows the direction the reaction actually proceeds in the")`
`color(blue)("cell above.")`

• If reduction is occurring at the `"Cu"` cathode (where we know the solution of `"Cu(NO3)"_2` ionizes to give off `"Cu"^(+2)` cations and `"NO"_3^(-1)` anions), this means `"Cu"^(+2)` ions will gain `2e^-` and will deposit onto the `"Cu"` electrode. The half reaction is the following for the left half cell:

`color(white)(aaaaaaaaaaaa)color(red)["Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s)`

`color(white)(aaaaaaaaaaaa)`

• In the right half cell, the `"Zn"` electrode (which is the anode) will lose `2e^-` and `"Zn"^(+2)` ions will fall out into the solution which already has `"Zn"^(+2)` cations and `"NO"_3^(-1)` ions from the zinc nitrate solution. The half reaction in the right cell is the following:

`color(white)(aaaaaaaaaaa)color(red)["Zn"(s)-> "Zn"^(+2)(aq) + 2e^(-)`

`color(blue)("C: Write a balanced chemical equation describing the total reaction of the")`
`color(blue)("electrochemical cell.")`

• If we combine the two half reactions, we get

`"Cu"^(+2)(aq) + cancel(2e^(-)) ->"Cu"(s)`
`"Zn"(s)-> "Zn"^(+2)(aq) + cancel(2e^(-))`
`--------------`
`color(red)["Cu"^(+2)(aq) + "Zn"(s)->"Cu"(s) + "Zn"^(+2)(aq)`

`color(blue)("D: What is the EMF of the cell above(under standard state conditions)")`

• Okay. We have established that at the `"Cu"` half cell, reduction will occur and at the `"Zn"` half cell. oxidation will occur. To find the EMF of the cell, the following equation is going to be used.

`color(white)(aaaaaaa)"E"^@("cell") = "E"^@("red") + "E"^@("oxidation")`

Since we were provided `"E"^@("red")` potentials, we have to reverse the reaction to show the oxidation of `"Zn"(s)` as well as reversing the value of the standard reduction potential to get the standard oxidation potential.

`"Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s)color(white)(aaaaa)"E"^@("red") = +0.34`

`"Zn"(s)->"Zn"^(+2)(aq) + 2e^(-)color(white)(aaaaa)"E"^@("oxidation") = +0.76`
`color(white)(aaaaa)`
- Use the equation to plug in and solve:

`color(white)(aaaaa)"E"^@("cell") = "E"^@("red") + "E"^@("oxidation")`

`color(white)(aaaaa)color(red)["E"^@("cell") = (+0.34) + (+0.76) = +1.1`